if 225 ml of 0.25M HCL and 25ml of 2.5M NaOH was used what is the limiting reagent in this experiment?
hi i don't know
I do these limiting reagent problems the long way.
HCl + NaOH ==> NaCl + H2O
mols HCl = M x L = about 0.056 but you need a better answer than that estimate.
That will produce 0.056 mols NaCl
mols NaOH = M x L = approx 0.0625.
That would produce approx 0.0625 mols NaCl.
In limiting reagent problems the smaller value ALWAYS wins and the reagent producing that value is the limiting reagent. That's HCl in this case.
To determine the limiting reagent in this experiment, we need to compare the number of moles of each reactant. The reactant that produces fewer moles will be the limiting reagent because it will be completely consumed, and any excess of the other reactant will not participate in the reaction.
First, let's find the number of moles of HCl:
Molarity of HCl = 0.25 M
Volume of HCl = 225 ml = 0.225 L
Using the equation: Moles = Molarity × Volume
Moles of HCl = 0.25 M × 0.225 L = 0.05625 mol
Next, let's find the number of moles of NaOH:
Molarity of NaOH = 2.5 M
Volume of NaOH = 25 ml = 0.025 L
Moles of NaOH = 2.5 M × 0.025 L = 0.0625 mol
Now, we compare the number of moles of HCl and NaOH. The mole ratio between HCl and NaOH is 1:1, so the reactant that produces fewer moles will be the limiting reagent.
In this case, HCl produces 0.05625 mol, while NaOH produces 0.0625 mol. Therefore, the limiting reagent is HCl because it produces fewer moles compared to NaOH.